Let
$$ K_m(f;x)=(-1)^{m(m-1)/2} \det[f^{(i+j)}(x)]_{i,j=0}^{m-1}. $$
If \(f\) has positive leading coefficient, degree at most six, and all of its zeros lie in \((-\infty,0]\), then
$$ K_m(f;x)\ge 0 $$
for every rank \(m\ge1\) and every \(x\ge0\).
Degree seven is the first place this can fail. Not merely the first degree where the proof breaks: there is an exact degree-seven counterexample at rank six, and it can be chosen with seven simple strictly negative zeros.
This is an all-rank theorem whose proof is secretly finite. That is the part I like.
Ranks one through three become positive root sums#
Write
$$ f(t)=a\prod_{\nu=1}^d(t+r_\nu), \qquad a>0,\quad r_\nu\ge0, $$
and set \(\alpha_\nu=(x+r_\nu)^{-1}\). Direct expansion gives
$$ \frac{K_2(f;x)}{f(x)^2}=\sum_\nu\alpha_\nu^2 $$
and
$$ \frac{K_3(f;x)}{f(x)^3} =8\sum_{\mu<\nu}\alpha_\mu^3\alpha_\nu^3 +12\sum_{\lambda<\mu<\nu} \alpha_\lambda^2\alpha_\mu^2\alpha_\nu^2. $$
There is no inequality trick left in these formulas. Every monomial is visibly nonnegative.
Ranks four, five, and six are supplied by three sharp finite-degree determinant theorems. Their respective degree ranges contain degree six, so the middle of the proof is literally a three-item checklist.
Rank seven has only one permutation#
Now suppose \(\deg f=6\). In the \(7\times7\) matrix
$$ [f^{(i+j)}(x)]_{i,j=0}^6, $$
every entry with \(i+j>6\) vanishes. A determinant permutation \(\sigma\) can contribute only if
$$ i+\sigma(i)\le6 $$
for every \(i\). But summing over \(i=0,\ldots,6\) gives \(42\) on both sides, so every inequality must be equality. The only surviving permutation is
$$ \sigma(i)=6-i. $$
Therefore
$$ K_7(f;x)=(6!a)^7>0. $$
This is extremely satisfying: the apparently complicated seventh-rank determinant has one living term.
For \(m\ge8\), the last column begins with a derivative of order at least seven, so it is identically zero. Every higher determinant vanishes. The infinite tail of ranks is finished by one dead column.
Why seven is genuinely sharp#
At degree seven, rank six can already be negative. A boundary example is
$$ f(x)=x^4(x+1)^3, \qquad K_6(f;0)=-1246946918400. $$
The verifier also checks a rational perturbation with seven distinct positive reciprocal roots, giving a polynomial with simple strictly negative zeros and the same negative sign. So repeated roots are not causing the failure.
The exact boundary matters. This theorem is about the signed centered consecutive-derivative family \(K_m\). It does not assert positivity of arbitrary Toeplitz minors, nonconsecutive Wronskians, or every determinant notion that happens to contain derivatives. But for this complete all-rank family, degree six is safe and degree seven is not. That edge is exact.
Notebook references: C-0065, C-0043, C-0056, C-0059